Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $t = \dfrac{3q + 15}{q + 9} \div \dfrac{-2q^2 - 12q + 54}{-q^2 - 6q + 27} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{3q + 15}{q + 9} \times \dfrac{-q^2 - 6q + 27}{-2q^2 - 12q + 54} $ First factor out any common factors. $t = \dfrac{3(q + 5)}{q + 9} \times \dfrac{-(q^2 + 6q - 27)}{-2(q^2 + 6q - 27)} $ Then factor the quadratic expressions. $t = \dfrac {3(q + 5)} {q + 9} \times \dfrac {-(q - 3)(q + 9)} {-2(q - 3)(q + 9)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {3(q + 5) \times -(q - 3)(q + 9) } {(q + 9) \times -2(q - 3)(q + 9) } $ $t = \dfrac {-3(q - 3)(q + 9)(q + 5)} {-2(q - 3)(q + 9)(q + 9)} $ Notice that $(q - 3)$ and $(q + 9)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-3\cancel{(q - 3)}(q + 9)(q + 5)} {-2\cancel{(q - 3)}(q + 9)(q + 9)} $ We are dividing by $q - 3$ , so $q - 3 \neq 0$ Therefore, $q \neq 3$ $t = \dfrac {-3\cancel{(q - 3)}\cancel{(q + 9)}(q + 5)} {-2\cancel{(q - 3)}(q + 9)\cancel{(q + 9)}} $ We are dividing by $q + 9$ , so $q + 9 \neq 0$ Therefore, $q \neq -9$ $t = \dfrac {-3(q + 5)} {-2(q + 9)} $ $ t = \dfrac{3(q + 5)}{2(q + 9)}; q \neq 3; q \neq -9 $